#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
#include <ctime>
using namespace std;
//黑科技最小堆实现方法
/*
priority_queue <int> q;
int main() {
//priority_queue comp() <int> q;
srand(time(0));
for (int i = 0; ; ++i) {
int n =rand()%10;
q.push(-1 * n);
cout << n <<" ";
if(i>10)break;
}
cout <<endl<<"dui"<<endl;
while (!q.empty()){
cout << -1 * q.top() <<" ";
q.pop();
}
return 0;
}
*/
struct Node{
int i;//顶点编号
int length;//当前路长
friend bool operator<(Node a, Node b) {//不加friend 就会出错
return a.length > b.length;//极小堆,默认情况是极大堆.
}
};
priority_queue < Node > q;
int c[10][10];//图的临街矩阵
int n;
const int INF = 1 << 30;
int dist[10];
int pre[10];
void graphShortestPath(int v){
Node E;
E.i = v;
E.length = 0;
dist[v] = 0;
q.push(E);//一开始没把初始点放进去,导致直接把第二个点pop掉了,没经过第二个点松弛边,所以一直出错.
while (1){
for (int j = 1; j <= n; ++j) {
if ( (c[E.i][j] < INF ) && (E.length + c[E.i][j] < dist[j]) ){
dist[j] = c[E.i][j] + E.length;
pre[j] = E.i;
Node temp;
temp.i = j;//此处一开始没有及时更新
temp.length = dist[j];
//cout << "temp=" << temp.i <<" "<< endl;
q.push(temp);
}
}
// try {
// q.pop();
// }
// catch (...) { break;}
if (q.empty())break;
q.pop();
E = q.top();
//cout << "q.top=" << q.top().i <<" ";
}
}
int main(){
int m;
//cin >> n >> m;
n = 6;
m = 9;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i == j) c[i][j] = 0;
else c[i][j] = INF;
}
}
for (int k = 1; k <= m; ++k) {
int t1,t2,t3;
cin >> t1 >> t2 >>t3;
c[t1][t2] = t3;
}
for (int l = 0; l < 10; ++l) {
dist[l] = INF;
pre[l] = 0;
}
graphShortestPath(1);
for (int i1 = 1; i1 <= n; ++i1) {
cout << dist[i1] <<" " ;
}
return 0;
}
/*测试数据:
1 2 1
1 3 12
2 3 9
2 4 3
3 5 5
4 3 4
4 5 13
4 6 15
5 6 4
运行结果:
0 1 8 4 13 17
*/
//测试ppriority_queue
/*
int main(){
Node temp1,temp2;
temp1.i=1;
temp1.length=11;
temp2.i = 2;
temp2.length = 22;
q.push(temp1);
q.push(temp2);
cout << q.top().i <<endl;
}*/
分支界限法实现单源最短路径
原文作者:分支限界法
原文地址: https://blog.csdn.net/gavinloverqq/article/details/51376362
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
原文地址: https://blog.csdn.net/gavinloverqq/article/details/51376362
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。